If u and v are integrable function of x, then, show that `intu.v.dx=uintvdx-int[(du)/(dx)intvdx]dx`. Hence evaluate `intlogxdx`.

Explanation

Let `intv.dx=w" "......(i)`
Integrating on both sides
`:.(dw)/(dx)=v`
using the chain rule for the derivatvie of the product of two functions.
`(d)/(dx)(u.w)=u.(dw)/(dx)+w.(du)/(dx)`
`=u.v+intv.dx.(du)/(dx)" "` [using (i)]
`=u.v+(du)/(dx).intv.dx`
Integrating both sides w.r.t. 'x', we get
`u.w=intu.vdx+int[(du)/(dx).intv.dx].dx`
`:'u=intv.dx`
`:.intu.vdx=uintv.dx-int[(du)/(dx).intv.dx].dx`
Henec Proved.
Now `intlogx.dx=intlogx.1.dx`
`=logx.int1.dx-int[(d)/(dx)(logx).int1.dx].dx`
`=logx.x-int(1)/(x).x.dx`
`=logx.x-int1.dx`
`=logx.x-x+C`
`=x(logx-1)+C`